How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x) = x / (3x-1)#?

1 Answer
Aug 16, 2016

Answer:

vertical asymptote at #x=1/3#
horizontal asymptote at #y=1/3#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: #3x-1=0rArrx=1/3" is the asymptote"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by x.

#(x/x)/((3x)/x-1/x)=1/(3-1/x)#

as #xto+-oo,f(x)to1/(3-0)#

#rArry=1/3" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1) Hence there are no oblique asymptotes.
graph{(x)/(3x-1) [-10, 10, -5, 5]}